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Boolean Algebra in Digital Systems: Multiplying Out, Factoring, and Simplification, Lecture notes of Digital Systems Design

PCI CollegeDigital Systems Design

In class notes and lecture notes from professor.

Typology: Lecture notes

2011/2012

Uploaded on 02/12/2018

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EEE 3342C: Digital Systems
Chapter 3: Boolean Algebra
(continued)
This chapter in the book includes:
3.1 Multiplying Out and Factoring Expressions
3.2 Exclusive-OR and Equivalence Operations
3.3 The Consensus Theorem
3.4 Algebraic Simplification of Switching Expressions
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Download Boolean Algebra in Digital Systems: Multiplying Out, Factoring, and Simplification and more Lecture notes Digital Systems Design in PDF only on Docsity!

EEE 3342C: Digital Systems

Chapter 3: Boolean Algebra

(continued)

This chapter in the book includes:

3.1 Multiplying Out and Factoring Expressions 3.2 Exclusive-OR and Equivalence Operations 3.3 The Consensus Theorem 3.4 Algebraic Simplification of Switching Expressions

Multiplying Out and Factoring Using Theorems

  • We saw before how to multiply out
    • Keep multiplying until there are no parentheses left
  • We saw before factoring
    • Keep factoring until everything is in the parentheses
  • Given an expression in product-of-sums form, the corresponding sum-of-products expression can be obtained by multiplying out, using the two distributive laws: X(Y + Z) = XY + XZ (3-1) (X + Y)(X + Z) = X + YZ (3-2)

Example

  • Multiply out this expression (Q + AB’).(C’D + Q’)
  • You can multiply out all the terms
  • But if you use the theorem in the box above, you end up with a simplified form It is equal to: QC’D + Q’AB’ (A+B).(A’+C) = AC + A’B

Example 2

  • Multiply out this expression
  • We are going to use the theorems in the boxes (A+B+C’).(A+B+D).(A+B+E).(A+D’+E).(A’+C) = (A+B +C’D).(A+B+E).(A+D’+E).(A’+C) = (A+B+C’DE).(A+D’+E).(A’+C) = (A+B+C’DE).[AC + A’(D’+E)] = (A+B+C’DE).(AC+A’D’+A’E) = AC + ABC + A’BD’ + A’BE + A’C’DE =AC + A’BD’ + A’BE + A’C’DE 5 (X+Y)(X+Z) = X + YZ (X+Y)(X’+Z) = XZ + X’Y Apply rule # X=A+B, Y=C’, Z=D Rule # Rule # Apply rule # X=A+B, Y=C’D, Z=E Apply rule # X=A, Y=D’+E, Z=C ABC disappears AC+ABC = AC.(1+B) = AC Note : if we were to multiply out by brute force, we would generate 162 terms, and 158 of these terms would then have to be eliminated to simplify the expression

Also called, the equivalence operation, A B A B 0 1 1 0 Exclusive OR (XOR) and Equivalence A B 0 0 0 1 1 0 1 1

A+B

A xnor B 1 0 0 1 Also written, A XOR B It is called equivalence operation because it is equal to 1 when A=B A B = A’B + AB’ it’s equal to 1 at terms 01 (A’B) and 10 (AB’) A B = AB + A’B’ it’s equal to 1 at terms 00 (A’B’) and 11 (AB)

Exclusive OR (XOR)

  • This is the symbol for XOR
  • Exclusive NOR and equivalence are the same thing

Equivalence Theorems (0 0) = 1 (0 1) = 0 (1 0) = 0 (1 1) = 1 (X Y) = X’Y’ + XY

Show by Boolean algebra that XOR and equivalence are the complement of each other (X Y)’ = (X’Y + XY’)’ = (X+Y’).(X’+Y) = XY + X’Y’ = (X Y)

Example

Apply DeMorgan’s (A+B)’ = A’B’, (AB)’ = A’+B’ Apply (A+B)(A’+C) = AC + A’B Here, A=X, B=Y’, C=Y

Example 3 Simplify this expression F = A’ B C F = [(A’)’B + A’B’] C = (AB + A’B’) C = (AB + A’B’)’C + (AB + A’B’)C’ = (A’B + AB’)C + (AB + A’B’)C’ = A’BC + AB’C + ABC’ + A’B’C’ X Y = X’Y + XY’ AB + A’B’ is the complement of A’B + AB’ The first one is ‘equivalence’ or XNOR, the second one is XOR

Consensus Theorem XY + X’Z + YZ = XY + X’Z (3-20) Proof: XY + X’Z + YZ = XY + X’Z + (X+X’)YZ = XY + X’Z + XYZ + X’YZ = XY(1+Z) + X’Z(1+Y) = XY + X’Z Dual Form: (X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z) (3-21)

Last Part from Exercises

  • Using theorems, factor the following expression G = A + BCD + D (E + F) = (A + BCD + D) (A + BCD + E +F) = [A + D(BC + 1)] (A + BCD + E + F) = (A + D) (A + E + F + B) (A + E + F + C) (A + E + F + D) W (W + V) = W + VW = W(1 + V) = W G = (A + D) (A + E + F + B) (A + E + F + C) X + YZ = (X + Y)(X + Z) X Y Z W = A + D V = E + F This is the last part of the exercises on Chapter 2 & 3.

Simplifying Boolean Expressions

1. Combining terms. Use the theorem XY + XY′ = X to combine two terms. 2. Eliminating terms. Use the theorem X + XY = X to eliminate redundant terms if possible; then try to apply the consensus theorem ( XY + X′Z + YZ = XY + X′Z ) to eliminate any consensus terms. abc′d′ + abcd′ = abd′ [X = abd′, Y = c] (3-24) Section 3.4 (p. 68-69) a′b + a′bc = a′b [X = a′b] a′bc′ + bcd + a′bd = a′bc′ + bcd [X = c, Y = bd, Z = a′b] (3-24)

Simplifying Boolean Expressions

4. Adding redundant terms. Chose to combine or eliminate other terms , ex. Add YZ to XY+XZ or XY to X WX + XY + X′Z′ + WY′Z′ (add WZ′ by consensus theorem) = WX + XY + X′Z′ + WY′Z′ + WZ′ (eliminate WY′Z′) = WX + XY + X′Z′ + WZ′ (eliminate WZ′) = WX + XY + X′Z′ (3-27)

Example (consensus Theorem)

  • Simplify the following logic function using the consensus theorem. F = ABCD + B’CDE + A’B’ + BCE’ +ACDE

F = B’CDE + A’B’ + BCE’ +ACDE

F = A’B’ + BCE’ +ACDE

XY + X’Z + YZ = XY + X’Z